A and B are variable points on X and Y axis respectively,such that l(ab)=4.Find the locus of the point which divides segment AB internally in the ratio 1:2.
I think that it must be a circle . or another curve it cant be a straight line.
[Math] Find the locus of the point which divides segment AB internally in the ratio 1:2.
analytic geometrycoordinate systemslocus
Related Solutions
There is a rather elegant solution to this, but requires some construction, and would be difficult to include in an answer, but I shall try my best.
Now, in the above diagram, $P$ is the moving point, and $C$ replaces $A$ in your question (sorry about that, software is new to me). When you deflect $P$ w.r.t to $CP'$ by $\theta$, the length of $CP = 2r\cos\theta$
Now, we know that $BQ = k.CP$ from the condition of our problem, hence
$$BQ = 2kr \cos \theta$$ $$BQ' = 2kr$$
Hence, we will always have
$$BQ = BQ' \cos \theta$$
And since $Q'$ is a fixed point for a given $C$ and $B$, this would be a circle. Actually a pair, because it could be on either side of $BP$. I hope this answers your question
Best Answer
Hint: It is a ellipse. To see this suppose that you have $A=(x,0)$ and $B=(0,y)$, so: $$ x^2+y^2=4 $$ Also there are tow points that divide the segment $AB$ internally by $1:2$ and these are $(s,t)=(\frac{x}{3},\frac{2y}{3})$ and $(z,w)=(\frac{2x}{3},\frac{y}{3})$. Now you can replace $x,y$ in the previous equation by $s,t,z,w$: $$ x^2+y^2=9s^2+\frac{9t^2}{4}=4 $$ and $$ x^2+y^2=\frac{9z^2}{4}+9w^2=4 $$ which are two ellipses.